After a brief neutron irradiation of silver, two isotopes are present: ${}^{\mathbf{108}}\mathbf{Ag}\mathbf{(}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{42}\mathbf{min}\mathbf{)}$with an initial decay rate of $\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{/}\mathit{s}$,and role="math" localid="1661598035621" ${}^{\mathbf{110}}\mathbf{Ag}\mathbf{(}{\mathbf{T}}_{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{=}\mathbf{24}\mathbf{.}\mathbf{6}\mathbf{s}\mathbf{)}$with an initial decay rate of. Make a semilog plot similar to Fig. 42-9 showing the total combined decay rate of the two isotopes as a function of time t = 0 from until t = 10min .We used Fig. 42-9 to illustrate the extraction of the half-life for simple (one isotope) decays. Given only your plot of total decay rate for the two-isotope system here, suggest a way to analyze it in order to find the half-lives of both isotopes.

1. The half life of ${}^{108}\mathrm{Ag},{\left({\mathrm{T}}_{1/2}\right)}_{108}=2.42\min \mathrm{or}145.2\mathrm{s}$
2. Decay rate of${}^{108}\mathrm{Ag},{\left(\mathrm{R}\right)}_{108}=3.1\times {10}^5/\mathrm{s}$
3. The half life of${}^{110}\mathrm{Ag},{\left({\mathrm{T}}_{1/2}\right)}_{110}=24.6\mathrm{s}$
4. Decay rate of${}^{110}\mathrm{Ag},{\left(R\right)}_{110}=4.1\times {10}^6\mathrm{s}$

The total combined decay rate of two-isotope system is as follows:

$InR=In\left(R_0{e}^{-\lambda t}+R_0\text{'}{e}^{-\lambda t}\right)$ …… (i)

The disintegration constant is as follows:

$\mathrm{\lambda }=\frac{\mathrm{In}2}{{\mathrm{T}}_{{\displaystyle \frac{1}{2}}}}$ …… (ii)

Here, ${\mathrm{T}}_{\frac{1}{2}}$is the half-life of the substance.

From the given data ${\mathrm{R}}_0=3.1\times {10}^5/\mathrm{s}$and${\mathrm{R}}_0=4.1\times {10}^6/\mathrm{s}$equation (i), the combined decay rate of the isotopes, the plot is made accordingly for disintegration constants using equation (ii) as:

$$\begin{gathered} \mathrm{\lambda }=\frac{\mathrm{In}2}{145.2\mathrm{s}} \\ \mathrm{\lambda }=\frac{\mathrm{In}2}{24.6\mathrm{s}} \end{gathered}$$.

The plot is given below:

Note that the magnitude of the slope for smallis$\lambda \text{'}$(the disintegration constant for${}^{110}\mathrm{Ag}$ ) , and for large tis $\lambda$(the disintegration constant forrole="math" localid="1661598893886" ${}^{108}\mathrm{Ag}$ ).